'When Are Relations Neither True Nor False?
in [Math]
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From: Newberry on 1 Apr 2010 00:49 On Mar 31, 7:30 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Newberry <newberr...(a)gmail.com> writes: > > No. > > I see. So you agree that for any formal theory T which is an extension > of Robinson arithmetic, either directly or through an interpretation, > and in which we can express statements of the form "the Diophantine > equation D(x1, ..., xn) = 0 has no solutions", there are infinitely many > true statements (of the form "the Diophantine equation D(x1, ..., xn) = > 0 has no solutions") that are unprovable in T if T is consistent? > Let me ask you something. Let ~(Ex)(Ey)(Pxy & Qy) (1) be Goedel's sentence. [Pxy: x is a proof of y, Qy is such that only one y=m satisfies it & m = #(1)]. You are absolutely convinced that PA is consistent i.e. (1) is true. That is, the search for x and y will never terminate. How do you know that the search for x and y will never terminate? > > You never answered my question what you ment by "Goedel." > > You have asked me what I mean by "Goedel"? > > -- > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > "Wovon man nicht sprechan kann, darüber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Daryl McCullough on 1 Apr 2010 06:09 Newberry says... >Let me ask you something. Let > >~(Ex)(Ey)(Pxy & Qy) (1) > >be Goedel's sentence. [Pxy: x is a proof of y, Qy is such that only >one y=m satisfies it & m = #(1)]. You are absolutely convinced that PA >is consistent i.e. (1) is true. That is, the search for x and y will >never terminate. How do you know that the search for x and y will >never terminate? Suppose you find two natural numbers m and n such that, P(n,m) & Q(m). Then you can easily prove (Ex) (Ey) (Pxy & Qy). You can also "decode" n to get a proof of ~(Ex)(Ey)(Pxy & Qy). So you would, in that case, have a proof of a contradiction. If your axioms are consistent, the the above case cannot happen. To say that "the above case cannot happen" is to say that there are no natural numbers m and n such that P(n,m) & Q(m). Which is formalized as ~(Ex)(Ey)(Pxy & Qy). So the assumption that your system is consistent directly leads to conclusion (1). -- Daryl McCullough Ithaca, NY
From: Aatu Koskensilta on 1 Apr 2010 08:59 Newberry <newberryxy(a)gmail.com> writes: > Let > > ~(Ex)(Ey)(Pxy & Qy) (1) > > be Goedel's sentence. [Pxy: x is a proof of y, Qy is such that only > one y=m satisfies it & m = #(1)]. You are absolutely convinced that PA > is consistent i.e. (1) is true. That is, the search for x and y will > never terminate. How do you know that the search for x and y will > never terminate? How do I know that Peano arithmetic is consistent? I know it the way I know any mathematical theorem I have personally proved. Perhaps you'd now be willing to say whether you agree that for any formal theory T extending Robinson arithmetic, either directly or through an interpretation, there are infinitely many true statements (of the form "the Diophantine equation D(x1, ..., xn) = 0 has no solutions") which are unprovable in T if T is consistent? -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Jesse F. Hughes on 1 Apr 2010 09:05 Newberry <newberryxy(a)gmail.com> writes: > On Mar 31, 5:57 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> But that is certainly not the argument that was offered in the other >> thread. The statement >> >> (En)(Ea,b,c) P(a,b,c,n) -> (En)(Ea,b,c)( P(a,b,c,n) & Q(a,b,c) ) >> >> was proved independently of your assumption, and no one balked. Thus, >> my question remains: was the argument I gave invalid? > > It is valid but it makes a silent assumption. An assumption so silent that no one but you noticed it? >> A second question comes to mind: what happened to your imagination >> test? You've said that (because FLT is true) you cannot picture >> (E a,b,c,n)( P(a,b,c,n) & Q(a,b,c) ). Yet, once you assume (contrary >> to fact) that FLT is false, you *can* picture it? > > No. >> >> You can't picture a green round triangle, right? What if I say: >> assume a round triangle exists. Can you picture a green round >> triangle *then*?[1] > > No. > >> That is, does the act of making an assumption >> change your capacities for imagining stuff? > > No. > >> If not, then your >> argument above doesn't work. > > Why not. Assume there are such things as meaningless sentences. These > sentences are well formed. Well formed sentences can go through > syntactic transformations. A similar syntactic transformation proves ~ (En)(Ea,b,c)( P(a,b,c,n) & Q(a,b,c) ) (1) This syntactic transformation does not apparently involve any assumptions at all. It seems to me that you're coming close to advocating an unsound theory: one which proves (1), despite the fact that you've said that (1) is not true. In any case, I really think you ought to think these things through. Up until now, you seem to have assumed that statements like (1) are useless if they're vacuously true. Now you've seen several proofs involving statements like (1) -- proofs which entail that these statements *are* vacuously true! It seems to me that quick responses to this observation are unlikely to work. Why not take the time and figure out what you think counts as a proof and see where that takes you? -- Jesse F. Hughes .... one of the main causes of the fall of the Roman Empire was that, lacking zero, they had no way to indicate successful termination of their C programs. -- Robert Firth
From: Nam Nguyen on 1 Apr 2010 20:30
Aatu Koskensilta wrote: > > How do I know that Peano arithmetic is consistent? I know it the way I > know any mathematical theorem I have personally proved. So what you're saying is you just _intuit_ PA system be consistent, no more no less. Of course anyone else could intuit the other way too! |